Question: If $x \dagger y = 3x^{2}+y^{2}$ and $x \star y = xy+4x-y$, find $(-5 \star -3) \dagger -1$.
Answer: First, find $-5 \star -3$ $ -5 \star -3 = (-5)(-3)+(4)(-5)-(-3)$ $ \hphantom{-5 \star -3} = -2$ Now, find $-2 \dagger -1$ $ -2 \dagger -1 = 3(-2)^{2}+(-1)^{2}$ $ \hphantom{-2 \dagger -1} = 13$.